A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer

To solve the problem, we will define the variables and set up the equations based on the information given.

### Step 1: Define Variables
Let:
- \( d \) = distance covered by the train (in km)
- \( s \) = speed of the train (in km/h)
- \( t \) = time taken to cover the distance (in hours)

From the relationship between distance, speed, and time, we have:
\[ d = s \times t \]

### Step 2: Set Up the Equations
1. **First Condition**: If the train were 10 km/h faster, it would take 2 hours less:
\[ \frac{d}{s + 10} = t - 2 \]
Rearranging gives:
\[ d = (s + 10)(t - 2) \]

2. **Second Condition**: If the train were 10 km/h slower, it would take 3 hours more:
\[ \frac{d}{s - 10} = t + 3 \]
Rearranging gives:
\[ d = (s - 10)(t + 3) \]

### Step 3: Equate the Two Expressions for Distance
Since both expressions equal \( d \), we can set them equal to each other:
\[ (s + 10)(t - 2) = (s - 10)(t + 3) \]

### Step 4: Expand Both Sides
Expanding both sides:
\[ st - 2s + 10t - 20 = st + 3s - 10t - 30 \]

### Step 5: Simplify the Equation
Cancelling \( st \) from both sides:
\[ -2s + 10t - 20 = 3s - 10t - 30 \]
Rearranging gives:
\[ 10t + 10t - 20 + 30 = 3s + 2s \]
\[ 20t + 10 = 5s \]
Dividing through by 5:
\[ 4t + 2 = s \quad \text{(Equation 1)} \]

### Step 6: Substitute \( s \) in Terms of \( t \) Back into One of the Distance Equations
Using the first condition:
\[ d = s \times t \]
Substituting \( s \) from Equation 1:
\[ d = (4t + 2)t \]
\[ d = 4t^2 + 2t \quad \text{(Equation 2)} \]

### Step 7: Use the Second Condition to Form Another Equation
Using the second condition:
\[ d = (s - 10)(t + 3) \]
Substituting \( s \) from Equation 1:
\[ d = (4t + 2 - 10)(t + 3) \]
\[ d = (4t - 8)(t + 3) \]
Expanding gives:
\[ d = 4t^2 + 12t - 8t - 24 \]
\[ d = 4t^2 + 4t - 24 \quad \text{(Equation 3)} \]

### Step 8: Set Equations 2 and 3 Equal to Each Other
Since both equations equal \( d \):
\[ 4t^2 + 2t = 4t^2 + 4t - 24 \]
Cancelling \( 4t^2 \):
\[ 2t = 4t - 24 \]
Rearranging gives:
\[ 24 = 4t - 2t \]
\[ 24 = 2t \]
\[ t = 12 \text{ hours} \]

### Step 9: Find the Speed \( s \)
Substituting \( t \) back into Equation 1:
\[ s = 4(12) + 2 = 48 + 2 = 50 \text{ km/h} \]

### Step 10: Find the Distance \( d \)
Using \( d = s \times t \):
\[ d = 50 \times 12 = 600 \text{ km} \]

### Final Answer
The distance covered by the train is **600 km**.

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